Question

If $I=\int \frac{\log \left(\cos x\right)}{{\cos }^{2}x}dx$, then $I$ equals

• A. $\tan x\log \cos x+\tan x-x+C$
• B. $\tan x\log \cos x-\tan x-x^2+C$
• C. $\tan x\log \cos x-\cot x+x+C$
• D. $\tan x\log \cos x+\cot x-x+C$

Answer 1

Answer: (A)

$\tan x\log \cos x+\tan x-x+C$

$I=\int \frac{\log \left(\cos x\right)}{{\cos }^{2}x}dx=\int {\sec }^{2}x\log \left(\cos x\right)dx=\tan x\log \left(\cos x\right)+\int {\tan }^{2}xdx=\tan x\log \left(\cos x\right)+\int ({\sec }^{2}x-1)dx=\tan x\log \left(\cos x\right)+\tan x-x+C$